A particle moving along the x axis in simple harmonic motion starts
from its equilibrium position, the origin, at t = 0 and moves to the
right. The amplitude of its motion is 1.70 cm, and the frequency is
1.10 Hz. Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and π.)

Using the equations:

$$
x(t) = A \cos(\omega t + \Phi)
$$
$$
\omega = 2\pi f
$$

I get A = 1.7cm or 0.017m, and
$$
\omega = 6.91
$$

I know that t = 0, x = 0. Thus,

$$
0 = 0.017 \cos(\Phi )
$$

And therefore,

$$
\Phi = \pi / 2
$$

From all of this, it seems to me that the equation for position with respect to time should be:

$$
x = 0.017 \cos(6.91t + \pi/2)
$$

Am I doing something wrong, because the above is not getting checked as the right answer (it's an online homework)

1 Answer
1

The cosine has more than one zero. And the text specifies that the particle goes to the right (I assume that the x axis also goes to the right). Now in which direction does the cosine go at $\pi/2$? And where's another zero?